Limit to Infinity of Power of x by Exponential of -a x

Theorem

Let $k$ and $a$ be positive real numbers.

Then:

$\ds \lim_{x \mathop \to \infty} x^k e^{-a x} = 0$

Proof

By Power Series Expansion for Exponential Function, we have:

$\ds e^{a x} = \sum_{n \mathop = 0}^\infty \frac {\paren {a x}^n} {n!}$

Since for $x > 0$ each term in this sum is non-negative, we have:

$\ds e^{a x} \ge \frac {\paren {a x}^{\floor k + 1} } {\paren {\floor k + 1}!}$

for each $k$.

So, for each $x > 0$ we have:

$\ds 0 < e^{-a x} \le \frac {\paren {\floor k + 1}!} {\paren {a x}^{\floor k + 1} }$

So that:

$\ds 0 \le x^k e^{-a x} \le \frac 1 {a^{\floor k + 1} } x^{k - \floor k - 1} \paren {\floor k + 1}!$

From the definition of the floor function, we have:

$0 \le k - \floor k < 1$

so:

$k - \floor k - 1 < 0$

Hence by Limit to Infinity of Power:

$\ds \lim_{x \mathop \to \infty} \frac 1 {a^{\floor k + 1} } x^{k - \floor k - 1} \paren {\floor k + 1}! = 0$

So, by the Squeeze Theorem, we have:

$\ds \lim_{x \mathop \to \infty} x^k e^{-a x} = 0$

$\blacksquare$